Partitioning the positive integers into finitely many arithmetic progressionsCovering the primes by 3-term...
Partitioning the positive integers into finitely many arithmetic progressions
Covering the primes by 3-term APs ?Is a “non-analytic” proof of Dirichlet's theorem on primes known or possible?Finitely many arithmetic progressionsThe Green-Tao theorem and positive binary quadratic formsWhat is the shortest route to Roth's theorem?Arithmetic progressions in power sequencesWhat are the limits of the Erdős-Rankin method for covering intervals by arithmetic progressions?Intersection of two arithmetic progressionsAbout consecutive integers covered by arithmetic progressionsProduct of arithmetic progressions
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago
add a comment |
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
arithmetic-progression elementary-proofs
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
edited 6 hours ago
Rodrigo de Azevedo
1,8422719
1,8422719
asked 14 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
JohnJohn
194
asked 14 hours ago
JohnJohn
194
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago
add a comment |
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago
2
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
1
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
John is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324088%2fpartitioning-the-positive-integers-into-finitely-many-arithmetic-progressions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
edited 8 hours ago
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 12 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
49.6k5114230
add a comment |
add a comment |
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f324088%2fpartitioning-the-positive-integers-into-finitely-many-arithmetic-progressions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
13 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
12 hours ago
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term. for k from 1 to inf { S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf } S(1) is odd integers; S(2) is 2, 6, 10 - twice an odd integer; and continues for powers of 2. In all cases the initial term is the difference / 2.
$endgroup$
– Judah Greenblatt
6 hours ago